I can\'tfigure this one out without more information. With onlydensity and molec

Question

I can'tfigure this one out without more information. With onlydensity and molecular mass given, how do you complete the equation:osmotic pressure = MRT. The answers are given, but I can'teven figure out how to plug them back in to find the solution inreverse.  Need some help here.

An aqueous dextrose solution having a density of 1.04 g/cm3 freezesat -1.15?C. Find the osmotic pressure of this solution at 25?C.[Given: for water Kf = 1.86?C/m; molecular mass of dextrose =180.16 g/mol] is it 13.8,   14.1, 12.9. 15.1, or 120 atm? I can'tfigure this one out without more information. With onlydensity and molecular mass given, how do you complete the equation:osmotic pressure = MRT. The answers are given, but I can'teven figure out how to plug them back in to find the solution inreverse.  Need some help here.

An aqueous dextrose solution having a density of 1.04 g/cm3 freezesat -1.15?C. Find the osmotic pressure of this solution at 25?C.[Given: for water Kf = 1.86?C/m; molecular mass of dextrose =180.16 g/mol] is it 13.8,   14.1, 12.9. 15.1, or 120 atm?

Explanation / Answer

delT = 1.15 C = Kf* molality molality = 1.15 C / Kf = 1.15C /(1.86 C/m) = 0.618 m = 0.618 moldextrose / kg solvent 1 kg solvent has 1000g of water AND 0.618 mol dextrose 0.618 mol dextrose* (180.16 g/mol) = 111.33888 grams total grams: 1000 + 111.33888 = 1111.3388 grams total volume of solution: 1111.3388 g *(1 cm3/ 1.04 g) *(1 mL/1cm3)* (1L/1000mL) = 1.06859508 L So 0.618 moles dextrose in 1.0685 L molarity = 0.618 mol/1.0685 L = 0.578 mol/L Osmotic pressure = molarity *R*T Osmotic pressure = (0.578 mol/L) *(0.0821 L*atm/mol/K) *(298K) Osmotic pressure = 14.1 atm

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